Question

If a and b are non-zero numbers such that ${\left(a^2+b^2\right)}^3={\left(a^3+b^3\right)}^2\mathrm{then}\frac{a}{b}+\frac{b}{a}=$
(a) $\frac{2}{3}$ (b) $\frac{3}{2}$ (c) $\frac{5}{6}$ (d) $\frac{6}{5}$

Answer 1

Dear Student,

$$\begin{gathered} {\left(a^2+b^2\right)}^3={\left(a^3+b^3\right)}^2 \\ applyformula{\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^{2}and{\left(a+b\right)}^2=a^2+b^2+2ab \\ \Rightarrow {\left(a^2\right)}^3+{\left(b^2\right)}^3+3{\left(a^2\right)}^2{b}^2+3a^2{\left(b^2\right)}^2={\left(a^3\right)}^2+{\left(b^3\right)}^2+2a^3{b}^3 \\ \Rightarrow a^6+b^6+3a^4{b}^2+3a^2{b}^4=a^6+b^6+2a^3{b}^3 \\ \Rightarrow 3a^4{b}^2+3a^2{b}^4=2a^3{b}^3 \\ divideby3a^3{b}^3 \\ \Rightarrow \frac{a}{b}+\frac{b}{a}=\frac{2}{3} \\ henceansweris\frac{2}{3}optionA \end{gathered}$$
Regards,